PHP:为什么exec()不返回输出? [英] PHP: Why isn't exec() returning output?

问题描述

我正在编写一个PHP脚本，用于通过Linux shell命令ping检查网络连接，并使用PHP的exec()调用它:

<?php


// Bad IP domain for testing.
$domain_bad = "lksjdflksjdf.com";

$ip_address = $domain_bad;

exec("ping -c 1 $domain_bad", $output, $return_var);

var_dump($return_var);
echo "return_var is: $return_var" . "\n";
var_dump($output);


exit;
?>

我没有从ping $output中得到错误消息的输出，这是我期望的:

$ php try.php
ping: unknown host lksjdflksjdf.com
int(2)
return_var is: 2
array(0) {
}

如果该域是一个很好的域，例如yahoo.com，则$output具有ping数组中的输出.但是，如果出现诸如'ping: unknown host lksjdflksjdf.com'之类的错误，则不会将其返回到$output数组.

为什么会发生这种情况，并且有更好的方法来做到这一点?

解决方案

您应将stderr重定向到stdout.

为此，请像这样更改exec()调用:

exec("ping -c 1 $domain_bad 2>&1", $output, $return_var);

有关2>&1的更多信息，意味着此处. >

I'm writing a PHP script to be used to check for network connections with Linux shell command ping calling it with PHP's exec():

<?php


// Bad IP domain for testing.
$domain_bad = "lksjdflksjdf.com";

$ip_address = $domain_bad;

exec("ping -c 1 $domain_bad", $output, $return_var);

var_dump($return_var);
echo "return_var is: $return_var" . "\n";
var_dump($output);


exit;
?>

I'm not getting the output for the error message from ping in $output which is what I'm expecting:

$ php try.php
ping: unknown host lksjdflksjdf.com
int(2)
return_var is: 2
array(0) {
}

If the domain is a good domain, such as yahoo.com, then $output has the output from ping in an array. But if it's an error such as 'ping: unknown host lksjdflksjdf.com' it doesn't get returned to the $output array.

Why is this happening and is there a better method to do this?

解决方案

You should redirect stderr to stdout.

To do that, change your exec() call like this:

exec("ping -c 1 $domain_bad 2>&1", $output, $return_var);

More info about 2>&1 meaning here.

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